Vector Quantity: A physical quantity that has both magnitude and direction. Example: Displacement, velocity, acceleration, force, momentum.

Part C: Answer any 6 questions. Each question carries 3 marks.

Derive the expression for the centripetal acceleration of a body performing uniform circular motion.
Answer: Consider a particle moving in a circle of radius $r$ with uniform speed $v$. Let the particle be at point A at time $t$ and at point B at time $t+\Delta t$. The change in velocity vector $\Delta \vec{v} = \vec{v_2} - \vec{v_1}$. From vector diagram (isosceles triangle formed by $\vec{v_1}$, $\vec{v_2}$, and $\Delta \vec{v}$), and considering similar triangles (position vectors $\vec{r_1}$, $\vec{r_2}$, and $\Delta \vec{r}$), we have: $\frac{|\Delta \vec{v}|}{|\vec{v}|} = \frac{|\Delta \vec{r}|}{|\vec{r}|}$ $\frac{\Delta v}{v} = \frac{\Delta r}{r}$ $\Delta v = \frac{v}{r} \Delta r$ Centripetal acceleration $a_c = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \lim_{\Delta t \to 0} \frac{v}{r} \frac{\Delta r}{\Delta t}$ Since $\lim_{\Delta t \to 0} \frac{\Delta r}{\Delta t} = v$ (speed), $a_c = \frac{v}{r} \cdot v = \frac{v^2}{r}$. In terms of angular velocity $\omega$: $v = r\omega$, so $a_c = \frac{(r\omega)^2}{r} = r\omega^2$.
Prove that for a projectile, the maximum height $H = \frac{u^2 \sin^2 \theta}{2g}$.
Answer: Consider the vertical motion of the projectile. Initial vertical velocity $u_y = u \sin \theta$. Final vertical velocity at maximum height $v_y = 0$. Acceleration $a_y = -g$. Using the kinematic equation: $v_y^2 = u_y^2 + 2a_y H$ $0^2 = (u \sin \theta)^2 + 2(-g)H$ $0 = u^2 \sin^2 \theta - 2gH$ $2gH = u^2 \sin^2 \theta$ $H = \frac{u^2 \sin^2 \theta}{2g}$.
State and prove the law of conservation of mechanical energy for a freely falling body.
Answer: Law of Conservation of Mechanical Energy: For a conservative force (like gravity), the total mechanical energy (sum of kinetic and potential energy) of a system remains constant, provided no non-conservative forces (like air resistance) do work on it. $E = K + U = \text{constant}$. Proof for a freely falling body: Consider a body of mass $m$ falling freely under gravity. Let its initial height be $h$ and it falls to the ground (height 0). At height $h$ (Point A): Initial velocity $u = 0$ (assuming dropped from rest). Kinetic Energy $K_A = \frac{1}{2}mu^2 = 0$. Potential Energy $U_A = mgh$. Total Mechanical Energy $E_A = K_A + U_A = 0 + mgh = mgh$. At height $x$ (Point B) from the ground: Let its velocity be $v$. Using $v^2 = u^2 + 2as$: $v^2 = 0^2 + 2g(h-x) = 2g(h-x)$. Kinetic Energy $K_B = \frac{1}{2}mv^2 = \frac{1}{2}m(2g(h-x)) = mg(h-x)$. Potential Energy $U_B = mgx$. Total Mechanical Energy $E_B = K_B + U_B = mg(h-x) + mgx = mgh - mgx + mgx = mgh$. At the ground (Point C, $x=0$): Let its velocity be $V$. Using $V^2 = u^2 + 2as$: $V^2 = 0^2 + 2gh = 2gh$. Kinetic Energy $K_C = \frac{1}{2}mV^2 = \frac{1}{2}m(2gh) = mgh$. Potential Energy $U_C = mg(0) = 0$. Total Mechanical Energy $E_C = K_C + U_C = mgh + 0 = mgh$. Since $E_A = E_B = E_C = mgh$, the total mechanical energy is conserved during free fall.
Define escape velocity. Derive an expression for escape velocity.
Answer: Escape Velocity ($v_e$): It is the minimum velocity with which a body must be projected vertically upwards from the surface of a planet so that it escapes the gravitational field of the planet and never returns. Derivation: Consider a body of mass $m$ projected from the Earth's surf